Evidence using GAP N:=Number of groups of order at most 2000 = 49910529484 k:=Number of groups of order 2^10 = 49487365422 k/N = .99152154732929340606... So with probability >.99 a group of order at most 2000 has order 2^10. So who knows about subgroups of 2-groups? On Fri, May 4, 2018 at 5:20 PM, <rcs@xmission.com> wrote:
Perhaps most groups are of order 2^N? --Rich
------ Quoting James Propp <jamespropp@gmail.com>:
My intuition is that (a) most groups have at least one non-normal subgroup, and (b) where there's one non-normal subgroup, there are lots of them (since conjugations via elements of G map them all over the place). But that's just my gut feeling.
Jim
On Fri, May 4, 2018 at 4:38 PM, Allan Wechsler <acwacw@gmail.com> wrote:
This is just not intuitively obvious to me. There are very common classes
of integers, so that for any n in the class, the chance of a random subgroup of a random group of order n being normal is quite high. For example, in the class of prime numbers, the only groups of order n are cyclic, and the only (two) subgroups are normal. Now, the chance of picking a prime n gets small as n gets large, but it doesn't get small very fast. It seems plausible to me that there might be classes of nonzero density, that have a large proportion of normal subgroups.
On Fri, May 4, 2018 at 4:07 PM, James Propp <jamespropp@gmail.com> wrote:
Has anyone proved that normally subgroups aren't normal?
(That is, for large n, if you pick a group G of order n uniformly at random, and then pick a subgroup H of G uniformly at random, the probability that H is a normal subgroup of G goes to 0 as n goes to infinity.)
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